Answer:
99.87% of the store’s total delivery orders will be delivered to consumers with charge
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
If a pizza store’s policy is, "Orders delivered within one hour or they’re free!", what percentage of the store’s total delivery orders will be delivered to consumers with charge?
Within one hour, which is 60 minutes. So this is the pvalue of Z when X = 60.
has a pvalue of 0.9987
99.87% of the store’s total delivery orders will be delivered to consumers with charge
Answer:
4
Step-by-step explanation:
Given that :
Clients are interviewed in groups of 2 on the first day; meaning two persons at a time
Second day, clients are interviewed in groups of 4; meaning 4 persons at a time.
Therefore, if the same number of clients are to be interviewed on each day, the smallest number of clients that could be interviewed each day could be obtained by getting the Least Common Multiple of both numbers: 2 and 4
- - - - 2 - - - 4
2 - - - 1 - - - 2
2 - - - 1 - - - 1
Therefore, the Least common multiple is (2 * 2) = 4
Therefore, the smallest number of clients that could be interviewed each day is 4.
<span>measure of ∠EGF = 1/2( 180 - 50)
= 1/2(130)
= 65
</span><span>the measure of ∠CGF = 180 - 65
= 115</span>
What is the question?
I'm assuming it is to find the length and width.
+_= plus or minus
(X+36)
____________
| |
(X) | |
|____________|
X^2+36X-2040<0
X<-36+_(36^2-4*-2040)^(1/2)
-----------------------------------
2
X<-18+_2((591)^(1/2))
This is probably not what you wanted, sorry
To find the specification limit such that only 0.5% of the bulbs will not exceed this limit we proceed as follows;
From the z-table, a z-score of -2.57 cuts off 0.005 in the left tail; given the formula for z-score
(x-μ)/σ
we shall have:
(x-5000)/50=-2.57
solving for x we get:
x-5000=-128.5
x=-128.5+5000
x=4871.50
